Termination w.r.t. Q proof of TRS/SK904.28.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))
MEM₂(g₂(x, y), z) -> MEM₂(x, z)
F₂(x, g₂(y, z)) -> F₂(x, y)
++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)
MEM₂(x, max₁(x)) -> NULL₁(x)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))
MEM₂(g₂(x, y), z) -> MEM₂(x, z)
F₂(x, g₂(y, z)) -> F₂(x, y)
++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)
MEM₂(x, max₁(x)) -> NULL₁(x)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MAX₁(x₁)) = 3·x₁²
POL(g₂(x₁, x₂)) = 2 + 3·x₁ + x₁·x₂
POL(u) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MEM₂(g₂(x, y), z) -> MEM₂(x, z)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MEM₂(g₂(x, y), z) -> MEM₂(x, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MEM₂(x₁, x₂)) = 2·x₁ + 2·x₁·x₂
POL(g₂(x₁, x₂)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(++¹₂(x₁, x₂)) = 2·x₁·x₂ + 2·x₂
POL(g₂(x₁, x₂)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F₂(x, g₂(y, z)) -> F₂(x, y)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

F₂(x, g₂(y, z)) -> F₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = 2·x₁·x₂ + 2·x₂
POL(g₂(x₁, x₂)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.